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What exactly do people mean when they talk about the scale dependence of the effective potential ($V$)? I explain the motivation for my question (and hence my confusion) below. Please correct me as appropriate.

If one defines the effective potential as the non-derivative part ($p^2 \rightarrow 0$) of the 1-PI effective action, then something like the Callan-Symanzik equation will imply that $$\frac{d}{d \log \mu} V = 0 \implies \left[ \frac{\partial}{\partial \log \mu} + \beta_i \frac{\partial}{\partial \lambda_i} + \gamma \frac{\partial}{\partial \log \phi} \right] V = 0$$

If the value of the effective potential at any field value (and zero momentum) is scale independent, then shouldn't the vacuum be stable at all scales, if we know it's stable at some scale?

All that I've said above holds so long as perturbation theory has been used correctly.

I've also seen some people "RG-improve" the effective potential by resumming the leading logs across all loop order. But then, the moment you do a partial resummation across loop orders, is there any reason for the thing you calculate to be scale independent? It's also not clear what the physical interpretation of such a scale-dependent quantity should be -- so why should one take the appearance of another vacuum seriously -- after all, you don't see any such thing when you do your calculations at some "low" scale -- and physical observables (at zero momentum) better not be RG-scale dependent?

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    $\begingroup$ Stability is a statement of the minimum of the effective potential. The coupling constants in the potential do change with scale as can be worked out from the Callan-Symanzik equation. Consider the $\phi^4$ theory -- suppose the sign of the quadratic term changes sign. Then the $\phi=0$ minimum could become unstable. $\endgroup$
    – suresh
    Commented Sep 6, 2014 at 2:50
  • $\begingroup$ See arxiv.org/abs/1408.0287 around eq.2.8 $\endgroup$
    – TwoBs
    Commented Sep 6, 2014 at 7:08

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I suggest you to read https://arxiv.org/abs/1408.0287 for a nice discussion on the RG dependence of the effective potentials and the RG invariant statements that one can make. About your point, I think you are simply misunderstanding what people mean: they are calculating at what point in $\Phi$, not in $\mu$, the potential cross zero.

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  • $\begingroup$ Thanks for that link; I like that paper. I understand that when talking about stability, one is referring to the existence of another minimum of $V_\mu(\phi)$. If I understand your response correctly, do you mean to say that when people talk about the profile of the effective potential at scale $\mu$ they take into account the explicit $\mu$ dependence and the running couplings, but not the field strength renormalization ("Method 2" as stated on page 8 of that paper)? If so, is that because one works at zero external momentum? $\endgroup$
    – Siva
    Commented Sep 6, 2014 at 8:55
  • $\begingroup$ As far as I understand, the stability bound that people discuss for the potential are RG invariant statements since the potential at the minimum $V_{min}$ is clearly RG invariant. In other words, looking for the limit value of the coupling $\lambda_{*}$ (defined at some IR scale) that gives $V_{EW}=V_{min}$ seems meaningful. If one experimentally see $\lambda_{exp}> \lambda_{*}$ (as measured at the scale $\mu$), then our vacuum is unstable. Saying instead that instability occurs at a scale $\mu$ where $V$ turns negative (or more correctly, when $V(\mu)<V_{EW}$) is a RG-dependent statement. $\endgroup$
    – TwoBs
    Commented Sep 6, 2014 at 12:47
  • $\begingroup$ In the case of the Higgs boson the role of the coupling $\lambda$ is actually played by its mass $m_h$ (or the top mass or both). $\endgroup$
    – TwoBs
    Commented Sep 6, 2014 at 12:56

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